A first-order differential equation is a linear equation in the dependent variable \(y\) if it can be written in the form:
\(a_1(x) \cfrac{dy}{dx} + a_0(x) y = g(x)\)
The standard form of a linear equation is:
\(\cfrac{dy}{dx} + P(x) y = f(x)\)
If \(f(x) = 0\) the equation is said to be homogeneous and is separable. If the equation is non-homogeneous, we can find an integration factor, \(\mu (x)\) such that:
\(\cfrac{d}{dx} (\mu (x) y) = \mu (x) (\cfrac{dy}{dx} + P(x) y ) \)
\( \cfrac{d \mu}{dx} y + \mu (x) \cfrac{dy}{dx} = \mu (x) \cfrac{dy}{dx} + \mu (x) P(x) y \)
\(\cfrac{d \mu}{dx} y = \mu (x) P(x) y \)
\(\cfrac{d \mu}{dx} = \mu (x) P(x) \)
This equation is separable with solution:
\(\mu (x) = e^ {\int P(x)}\)
Find the integrating factor for the DE \(t y^{'} - 3 y = \cfrac{\sin {\pi t}}{t}\).
First, we need to write the DE in standard form:
\(y^{'} - \cfrac{3}{t} y = \cfrac{\sin {\pi t}}{t^2}\)
Where \(P(t) = - \cfrac{3}{t} \)
Next, we can determine the integrating factor:
\(\mu(t) = e^ {\int P(t)}\)
\(\mu(t) = e^ {\int - \frac{3}{t} dt}\)
\(\mu(t) = e^{-3 \ln t }\)
\(\mu(t) = t^{-3}\)
Therefore, we can determine the integrating factor is \(\boldsymbol{t^{-3}}\).
The steps to solve a linear non-homogeneous differential equation are:
Solve the DE \(t^3 y^{'} +4t^2 y = e^{-t} \) subject to \(y(-1) = 0 \).
First, write the DE in standard form:
\( y^{'} + \cfrac{4}{t} y = \cfrac{e^{-t}}{t^3} \)
Next, identify \(P(x)\) and find the integrating factor:
\(\mu (x) = e^ {\int P(x)}\)
\(\mu (t) = e^ {\int \cfrac{4}{t} dt}\)
\(\mu (t) = e^ {4 \ln t}\)
\(\mu (t) = t^4\)
Then, we can multiply both sides of the DE by the integration factor:
\( y^{'} + \cfrac{4}{t} y = \cfrac{e^{-t}}{t^3} \)
\(t^4 \! \left(y^{'} + \cfrac{4}{t} y\right) = t^4 \cfrac{e^{-t}}{t^3}\)
After, we can replace the left-hand side of the equation and simplify the right-hand side:
\(\cfrac{d}{dt}(t^4 y) = t e^{-t} \)
Notice the left-hand side, by chain rule is:
\(\cfrac{d}{dt}(t^4 y) = t^4 y^{'} + 4t^3y = t^4 \! \left(y^{'} +\cfrac{4}{t}y \right)\)
Next, we can take the integral of both sides:
\( \int{ \cfrac{d}{dt}(t^4 y) dt} = \int{ t e^{-t} dt} \)
The left hand side integral and derivative cancel out. The right hand side can be solved using integration by parts:
| \( u = t\) | \( u^{'} = 1\) |
| \( v = e^{-t}\) | \( \int v = -e^{-t}\) |
\( \int{ t e^{-t} dt} = -te^{-t} - \int {-e^{-t} dt} \)
\( \int{ t e^{-t} dt} = -te^{-t} - e^{-t} + c\)
\( \int{ \cfrac{d}{dt}(t^4 y) dt} = \int{ t e^{-t} dt} \)
\( t^4 y = -te^{-t} - e^{-t} + C \)
\(y = \cfrac{-e^{-t}}{t^3} - \cfrac{e^{-t}}{t^4} + \cfrac{C}{t^4} \)
Now we can solve for the constant using \(y(-1) = 0 \):
\(y(-1) = 0 = \cfrac{-e^{1}}{-1} - \cfrac{e^{1}}{1} + \cfrac{C}{1} \)
\(0 = C \)
\(y = \cfrac{-e^{-t}}{t^3} - \cfrac{e^{-t}}{t^4} \)
Finally, we can factor the expression further:
\(y = -(t+1)\cfrac{e^{-t}}{t^4}\)
Therefore, we can determine the solution to the DE is \(\boldsymbol{y = -(t+1)\cfrac{e^{-t}}{t^4}}\).